I think I've got a proof that angle BDE = 20 degrees. I'll try throwing some diagrams together in Paint Shop Pro. It uses properties of isosceles trapeziums, which may cause some confusion as it seems that Americans call trapeziums trapezoids, and conversely call trapezoids trapeziums!
I've tried to prove it cleanly by angle wrangling, but I'm getting bogged down trying to simplify a bazillion trig expressions and making silly errors..At some point I have to use the angle bisector theorem or the cosine rule...both of which yield ugly expressions that have only worked so far with my calculator... There must be a simple and elegant proof but I'm still searching for it 😳
Ok, you're right, it's 30 degrees. Here is my solution. which I think is resonably elegant.
1) Add another point, F, between D and C such than angle CBF = 20 degrees.
2) Angles in triangle BCF add up to 180, so BFC = 80.
3) Hence triangle BCF is isosceles, with BC = BF.
4) Angles in triangle BEC add up to 180, so BEC = 50.
5) Hence triangle BEC is isosceles, with BE = BC.
6) So BC = BF = BE.
7) Angle FBE is 80 - 20 = 60.
8) Hence triangle FBE has two equal sides and one angle is 60, thus it is equalateral.
9) Hence EF = BF (= BC = BE)
10) Angles in triangle BDC add up to 180, so angle BDC = 40.
11) Angle DBF also equals 40.
12) Hence triangle FBD is isosceles, with BF = DF.
13) So we have DF = EF
14) Hence triangle FED is isosceles, with EDF = FED = (180 - 40) / 2 = 70.
15) Hence BDE = 70 - BDC = 70 - 40 = 30.